Full text of "127"
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107
Assume the symbolism -\-y/j=— 1, j being an impossible quantity, the root
of the equation + |/a;-(-l=0.
Squared), l + 10/j/-3/y s =4;', y*(4/-l)-10y+3=0, from which
10j±, /(7-3;)
a- 1
III. Solution by the PROPOSER.
Transpose and square.
.•. 4x 8 — 20x+25— z s — 7 (B), an obvious quadratic.
Apply its roots, 4 and §, to the given {A); hence 2(4)+[— 3]=8— 3=5 ;
=2z+{-[,/(16-7)]} (C);and
2(|) + (-1 )=6i-i=5 =2*+ { -[,/ ( V - V)] } •
■(D);
satisfy it. Could extracting \/(x % — 7) positive here also yield roots, then M)'s
dominant quadratic (B) is bi-quadratic, which is absurd.
Also solved by P. S. BERG and CHAS. C. CR08S.
GEOMETRY.
127. Proposed by WILLIAM HOOVES. A. M., Ph. D., Professor of Mathematics and Astronomy, Ohio Uni-
versity, Athens. Ohio.
The equation to the plane through the extremities, (x,, j/,, «,), (x t , y s , z s ),
( x si 2/s , Sj)i of conjugate diameters of the ellipsoid,
lL+ _Vl_ + _£!__, is l\ ±*z ±^ r .VrtVi +y» v ■ «i+' t +«i ,._ 1
Solution by the PROPOSER.
If Za;+»M/+«2=p ... (1) be the required plane, we should have
te 1 +my 1 +7M! t =p (2),
Jx 2 + mj/ 2 + nz s =p (3),
tes + my,i+«*»=P (4).
Solving these for £/p, w/p, ?i/p, we have
l/prr.
1. »!.«»
If 3/2. «J
1, ?/.,. 2 3
^i i 2/it 2 i
**» 2^2 > Z 2
^3! 3/31 23
• (5).
w>/p=etc., n/p=etc (6).
Reducing (5), making use of
108
r, % A. 1 .2 ~2 r 2
■^ + "^- + ^~- 1 -^" + -Si" 1 " •■•• (?) '
and a; ,1/, +x. i y l +x. i y 3 :-y,z^+etc.,~ z,*, -t-etc.,=0 (8),
Z/p^-'-i^i-tii, m/p=etc, n/p=etc, (0).
These must be put into (1).
Also solved by G. B. M. ZBBB, J. W. YOUNG, LON C. WALKEB, J. SCBBFFEB, and GEOBOE
LILLEY.
128. Proposed by W. H. CARTER, Vice President and Professor of Mathematics. Centenary College, Jack-
son. La.
Given i i '=A n ~ 1 -j-(n— 1)!. A ,. A s a„, where a— -the determinant
(«,6 s c 3 k„) and A,A, A„ are the minors of the elements of the »th
column ; and a m , b m , c m : etc.. (m— 1, 2, 3 n) are the coefficients of n
given equations containing w—1 variables. Show (1) that n—S, F=the area of
a triangle, and (2) if «=4, F=the volume of the tetrahedron.
Solution by J. W. YOUUG, Student in Ohio State University, Columbus. 0.
1. Let n—'d. The points of intersection of the three lines represented by
the given equations, are
*.
B,
_ *,.
V\ — 7j j 2/2 — ^T~» 2/s jf i
where, by the usual notation, A k equals the co-factor a t , in the determinant A .
The area of the triangle formed by these points is
Area—}
*,,
3/1. 1
^2 •>
2/2. 1
=i
^31
2/3. 1
c.
1
_d? g 2
C, ' c 2
A 3 B 3
1
=»
^1.
B„
c,
A,,
Bu
c„
A*,
B 3 ,
c,
-*-<?, C 8 a
and this, by a well-known theorem in determinants.
=}A 8 -f-C,C 2 C , .,=F.
2. Let n—4. The intersections of the four planes given by the equations
are found precisely as above.